 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/discrete-mathematics-continuous-mathematics-and-probability/9/b/i/solution

μ > 0

. If a nonnegative random variable had zero expected value, it would equal zero almost surely, which is impossible for a distribution having a probability density function on

[0, \infty)

: a singleton has probability zero. Thus the given division by

μ

is legitimate.

Because

x \geq 0

and

f (x) \geq 0

g (x) = x f (x) / μ \geq 0

. Moreover

\int_{0}^{\infty} g (x) d x = \frac{1}{μ} \int_{0}^{\infty} x f (x) d x = \frac{μ}{μ} = 1.

Extend

g

by zero for negative

x

. It is a probability density function, defining the size-biased distribution of

X

: larger observations receive weight proportional to their value.

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